Q. The focal chord to $y^{2} = 16x$ is tangent to ${(1 - 6)}^{2} + y^{2} = 2$, then the possible values of the slope of this chord are
A) $\{ - 1,\ 1\}$
B) $\{ - 2,\ 2\}$
C) $\left\{ - 2,\ \frac{1}{2} \right\}$
D) $\left\{ 2, - \frac{1}{2} \right\}$
Explanation
(a)
Given equation of circle is ${(x - 6)}^{2} + y^{2} = \left( \sqrt{2} \right)^{2}$
[image:11_image83.emf]
$BC =$ radius $= \sqrt{2}$
The length of the tangent from $S$ to $B$
$\therefore SB = \sqrt{(4 - 6)^{2} + 0 - 2} = \sqrt{2^{2} - 2} = \sqrt{2}$
From figure, $\mathrm{\Delta}CBS$ is an isosceles triangle
$\Rightarrow \theta = 45{^\circ} \Rightarrow m = 1$ $(\because BC = BS)$
Similarly, for $\bigtriangleup CSD,\ m = - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.